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In the reaction: Na(2)S(2)O(3)+4Cl(2)+5H...

In the reaction: `Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl`, the equivalent weight of `Na_(2)S_(2)O_(3)` will be: (M= molecular weight of `Na_(2)S_(2)O_(3))`

A

M/4

B

M/8

C

M/1

D

M/2

Text Solution

Verified by Experts

The correct Answer is:
B
`Na_(2)overset(+2)S_(2)O_(3)rarrNa_(2)overset(+6)SO_(4)`
the total change in oxidation number `=4xx2=8`
`:. E_(Na_(2)S_(2)O_(3))=(mol.wt.)/(V.f)=(M)/(8)`
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